Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $a = \dfrac{y^2 - 5y}{y^2 - 25} \times \dfrac{-8y - 40}{-8y - 80} $
Solution: First factor the quadratic. $a = \dfrac{y^2 - 5y}{(y + 5)(y - 5)} \times \dfrac{-8y - 40}{-8y - 80} $ Then factor out any other terms. $a = \dfrac{y(y - 5)}{(y + 5)(y - 5)} \times \dfrac{-8(y + 5)}{-8(y + 10)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ y(y - 5) \times -8(y + 5) } { (y + 5)(y - 5) \times -8(y + 10) } $ $a = \dfrac{ -8y(y - 5)(y + 5)}{ -8(y + 5)(y - 5)(y + 10)} $ Notice that $(y - 5)$ and $(y + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -8y(y - 5)\cancel{(y + 5)}}{ -8\cancel{(y + 5)}(y - 5)(y + 10)} $ We are dividing by $y + 5$ , so $y + 5 \neq 0$ Therefore, $y \neq -5$ $a = \dfrac{ -8y\cancel{(y - 5)}\cancel{(y + 5)}}{ -8\cancel{(y + 5)}\cancel{(y - 5)}(y + 10)} $ We are dividing by $y - 5$ , so $y - 5 \neq 0$ Therefore, $y \neq 5$ $a = \dfrac{-8y}{-8(y + 10)} $ $a = \dfrac{y}{y + 10} ; \space y \neq -5 ; \space y \neq 5 $